Let $p=4x-7$. Which equation is equivalent to $(4x-7)^2+16=40x-70$ in terms of $p$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $p^2+10p+16=0$ (Choice B) B $p^2+10p+86=0$ (Choice C) C $p^2-10p+86=0$ (Choice D) D $p^2-10p+16=0$
Solution: We are asked to rewrite the equation in terms of $p$, where ${p}={4x-7}$. In order to do this, we need to find all of the places where the expression ${4x-7}$ shows up in the equation, and then substitute ${p}$ wherever we see them! For instance, note that $40x-70=10({4x-7})$. This means that we can rewrite the equation as: $(4x-7)^2+16=40x-70$ $({4x-7})^2+16=10({4x-7})$ [What if I don't see this factorization?] Now we can substitute ${p}={4x-7}$ : $({p})^2+16=10({p})$ Finally, let's manipulate this expression so that it shares the same form as the answer choices: ${p}^2-10{p}+16=0$ In conclusion, $p^2-10p+16=0$ is equivalent to the given equation when $p=4x-7$.